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3 years ago

Structure of tetrazine

Chemistry

1 answer:

Vika [28.1K]3 years ago

3 0

Answer:

Tetrazine is a compound that consists of a six-membered aromatic ring containing four nitrogen atoms with the molecular formula C2H2N4.

(See the image)

Hope it helps!

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A solution used to chlorinate a home swimming pool contains 7% chlorine by mass. An ideal chlorine level for the pool is one par

deff fn [24]

Answer:

Volume of chlorine = 61.943 mL

Explanation:

Given:

Volume of the water in the Pool = 18,000 gal

also,

1 gal = 3785.412 mL

thus,

Volume of water in pool = 18,000 × 3785.412 = 68,137,470 mL

Density of water = 1.00 g/mL

Therefore,

The mass of water in the pool = Volume × Density

The mass of water in the pool = 68,137,470 mL × 1.00 g/mL = 68,137,470 g

in terms of million = $\frac{\textup{68,137,470}}{\textup{ 1,000,000}}$

= 68.13747 g

also,

1 g of chlorine is present per million grams of water

thus,

chlorine present is 68.13747 g

Now,

volume = $\frac{\textup{Mass}}{\textup{Density}}$

Volume of chlorine = $\frac{\textup{68.13747 g}}{\textup{1.10 g/mL}}$

Volume of chlorine = 61.943 mL

3 0

3 years ago

Carbon-14 (14C) dating assumes that the carbon dioxide on the Earth today has the same radioactive content as it did centuries a

Nataly [62]

Answer: The tree was burned 16846.4 years ago to make the ancient charcoal

Explanation:

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half life of the reaction = 5715 years

Putting values in above equation, we get:

$k=\frac{0.693}{5715yrs}=1.21\times 10^{-4}yrs^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $1.21\times 10^{-4}yr^{-1}$

t = time taken for decay process = ? yr

$[A_o]$ = initial amount of the sample = 100 grams

[A] = amount left after decay process = 13 grams

Putting values in above equation, we get:

$1.21\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{13}\\\\t=16864.4yrs$

Hence, the tree was burned 16846.4 years ago to make the ancient charcoal

8 0

3 years ago

how much heat is released when 70.9g of water at 66C cools to 25C? The specific heat of water is 1 cal/gC

Serga [27]

Answer:

-12162.47 joules (or -12000 joules when accounting for significant figures)

Explanation (btw I used 1 cal as 4.184 joules because SI units are better):

q = m c delta T

q = (70.9) (4.184) (25 - 66)

q = (70.9) (4.184) (-41)

q = -12162.47 joules

7 0

3 years ago

In the reaction 2co ( g) + o2( g) → 2co2( g), what is the ratio of moles of oxygen used to moles of co2produced?

KIM [24]

In the reaction 2co ( g) + o2( g) → 2co2( g), the ratio of moles of oxygen used to moles of co2produced is 1:2.

8 0

3 years ago

CO2+2H2O=H3O^++HCO3^-

bekas [8.4K]

Answer:

Bronsted Lowry Theory

$CO_2$ is the acid

$H_2 O$ is the base

$H_3 O^+$ is the conjugate acid and

$HCO_3^-$ is the conjugate base

Explanation:

Bronsted Lowry Theory:

An acid is a substance that can donate one or more protons

A base is a substance which can accept one or more protons

Hydrogen atom which is neutral (No Charge) contains 1 positive proton, 1 negative electron and 0 neutral neutron.

Thus Hydrogen atom has no Charge and it is neutral.

When an hydrogen atom loses an electron, Hydrogen ion is formed, which will contain 1 positive proton and 0 negative electron and 0 neutral neutron.

Thus Hydrogen ion has a positive charge.

Hydrogen ion is also called as a proton since it has only 1 proton in it.

Hydrogen ion in water that is,

$H^++H_2O\Rightarrow H_3O^+$

$H_3O^+$ is called as Hydronium ion.

Acid loses H+ (proton) to form conjugate Base

Base gains H+ (proton) to form conjugate Acid

For example

Let us consider the example given in the question

$CO_2+2H_2 O>H_3 O^+ +HCO_3^-$ can be written as (removed 1 $H_2 O$ from both the sides )

$CO_2+H_2 O>H^++HCO_3^-$ or

$CO_2+H_2 O > H_2 CO_3$ reversing the equation

$H_2 CO_3+H_2 O>H_3 O^+ +HCO_3^-$

$H_2 CO_3$ is the acid which donates $H^+$ to form $HCO_3^-$

$H_2 O$ is the base which gains $H^+$ to form $H_3 O^+$

$HCO_3^-$ is the conjugate base and $H_3 O^+$ is the conjugate acid

$CO_2$ is the acid

$H_2 O$ is the base

$H_3 O^+$ is the conjugate acid and

$HCO_3^-$ is the conjugate base

(Answer)

5 0

3 years ago